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Moving rod viewed in different inertial frames

  1. Jun 26, 2016 #1
    1. The problem statement, all variables and given/known data
    Two inertial frames S and S' are in standard configuration, the frame S' is moving along the x axis of S with velocity v. In S' a straight rod parallel to the x' axis moves in the y' direction with velocity u. Show that in S the rod is inclined to the x-axis at an angle ##- tan^{-1}(\frac{γuv}{c^2})##

    2. Relevant equations
    ## t = γ(t' + \frac{vx'}{c^2}) ~~~ x = γ(x' + vt') ~~~ Lorentz~~Transformation##
    ## γ = \frac{1}{[1 - (\frac{v}{c})^2]^½} ~~~ Lorentz~~ Factor##

    3. The attempt at a solution
    From the relativity of simultaneity, the front and back end of the rod will not be seen by the observer in S as having the same position as seen by the observer in S', the front end will lag behind the back end since light from the front will travel a little bit longer than the back to reach S.
    I will treat the front and back end as separate case, let P1, P2 be the back and front end respectively. Place P1 at x' = 0 so that P2 is at x' = L

    As seen by S, P1 is
    ##t = γ( t' + 0 ) = γt' ~~⇒~~ t' = \frac{t}{γ}##
    ##x = γ( 0 + vt' ) = γvt' = vt##
    ##y = y' = ut' = \frac{ut}{γ}##
    ##P1 = ( vt, \frac{ut}{γ} )##

    As seen by S, P2 is
    ##t = γ( t' + \frac{vL}{c^2} )~~⇒~~ t' = \frac{t}{γ} - \frac{vL}{c^2}##
    ##x = γ( L + vt' ) = γL + γvt' = γL + γv( \frac{t}{γ} - \frac{vL}{c^2} ) = γL + vt - γL\frac{v^2}{c^2} = γL(1 - \frac{v^2}{c^2} ) + vt = \frac{L}{γ} + vt##
    ##y = y' = ut' = u( \frac{t}{γ} - \frac{vL}{c^2} ) = \frac{ut}{γ} - \frac{uvL}{c^2}##
    ##P2 = ( vt + \frac{L}{γ}, \frac{ut}{γ} - \frac{uvL}{c^2})##

    ##tan(θ) = \frac{Δy}{Δx} = \frac{ \frac{ut}{γ} - \frac{uvL}{c^2} - \frac{ut}{γ} }{ vt + \frac{L}{γ} - vt } = \frac{ - \frac{uvL}{c^2} }{ \frac{L}{γ} } = -γ\frac{uv}{c^2}##

    ##tan(-θ) = -tan(θ)~~~⇒~~~ tan(-θ) = γ\frac{uv}{c^2} ~~~⇒~~~ θ = -~ tan^{-1}(γ\frac{uv}{c^2})##

    Is my solution correct?
     
  2. jcsd
  3. Jun 27, 2016 #2
    yes, correct !
     
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